Key Concepts and Formulas

• Variable Separable Differential Equations: A differential equation of the form $f(y)dy = g(x)dx$ can be solved by integrating both sides.
• Trigonometric Identity: $\cos(2\theta) = 2\cos^2(\theta) - 1$, therefore $\cos(\theta) = 2\cos^2(\theta/2) - 1$, which means $\cos(\theta/2) = \sqrt{\frac{1+\cos\theta}{2}}$.
• Integration by Substitution: A technique used to simplify integrals by substituting a function of $x$ with a new variable.

Step-by-Step Solution

Step 1: Simplify the trigonometric term

We are given the differential equation: $\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy$ Let $\theta = \cos^{-1}(e^{-x})$. Then $\cos(\theta) = e^{-x}$. We want to find $\cos(\frac{\theta}{2})$. Using the half-angle formula: $\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos(\theta)}{2}} = \sqrt{\frac{1 + e^{-x}}{2}}$ So, the differential equation becomes: $\sqrt{\frac{1 + e^{-x}}{2}} dx = \sqrt{e^{2x} - 1} dy$

Step 2: Separate variables

We want to separate the variables $x$ and $y$. Divide both sides by $\sqrt{e^{2x}-1}$ and multiply by $dy$, and divide both sides by $\sqrt{\frac{1+e^{-x}}{2}}$ and multiply by $dx$: $\frac{dy}{\sqrt{\frac{1 + e^{-x}}{2}}} = \frac{dx}{\sqrt{e^{2x} - 1}}$ $dy = \sqrt{\frac{1 + e^{-x}}{2}} \cdot \frac{dx}{\sqrt{e^{2x} - 1}} = \frac{1}{\sqrt{2}} \sqrt{\frac{1 + e^{-x}}{e^{2x} - 1}} dx$ $dy = \frac{1}{\sqrt{2}} \sqrt{\frac{1 + e^{-x}}{(e^x - 1)(e^x + 1)}} dx = \frac{1}{\sqrt{2}} \sqrt{\frac{e^x + 1}{e^x(e^x - 1)(e^x + 1)}} dx$ $dy = \frac{1}{\sqrt{2}} \sqrt{\frac{1}{e^x(e^x - 1)}} dx = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{e^x(e^x - 1)}} dx = \frac{1}{\sqrt{2}} \frac{dx}{\sqrt{e^{2x} - e^x}}$

Step 3: Integrate both sides

$\int dy = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{e^{2x} - e^x}}$ $y = \frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{e^{2x} - e^x}} + C$ Let $u = e^x$. Then $du = e^x dx$, so $dx = \frac{du}{u}$. $y = \frac{1}{\sqrt{2}} \int \frac{du}{u\sqrt{u^2 - u}} + C = \frac{1}{\sqrt{2}} \int \frac{du}{u\sqrt{u(u - 1)}} = \frac{1}{\sqrt{2}} \int \frac{du}{u\sqrt{u}\sqrt{u - 1}}$ Let $v = \sqrt{u-1}$. Then $v^2 = u-1$, so $u = v^2 + 1$. $du = 2v \, dv$. $y = \frac{1}{\sqrt{2}} \int \frac{2v \, dv}{(v^2 + 1)\sqrt{v^2 + 1}v} + C = \frac{2}{\sqrt{2}} \int \frac{dv}{(v^2 + 1)^{3/2}} + C$ Let $v = \tan \theta$. Then $dv = \sec^2 \theta \, d\theta$. $y = \sqrt{2} \int \frac{\sec^2 \theta \, d\theta}{(\tan^2 \theta + 1)^{3/2}} + C = \sqrt{2} \int \frac{\sec^2 \theta \, d\theta}{(\sec^2 \theta)^{3/2}} + C = \sqrt{2} \int \frac{\sec^2 \theta}{\sec^3 \theta} d\theta + C$ $y = \sqrt{2} \int \frac{1}{\sec \theta} d\theta + C = \sqrt{2} \int \cos \theta \, d\theta + C = \sqrt{2} \sin \theta + C$ Since $v = \tan \theta$, $\sin \theta = \frac{v}{\sqrt{v^2 + 1}}$. $y = \sqrt{2} \frac{v}{\sqrt{v^2 + 1}} + C = \sqrt{2} \frac{\sqrt{u - 1}}{\sqrt{u}} + C = \sqrt{2} \frac{\sqrt{e^x - 1}}{\sqrt{e^x}} + C = \sqrt{2} \sqrt{1 - e^{-x}} + C$

Step 4: Apply the initial condition

We are given that the curve intersects the y-axis at $y = -1$. This means when $x = 0$, $y = -1$. $-1 = \sqrt{2} \sqrt{1 - e^0} + C = \sqrt{2} \sqrt{1 - 1} + C = 0 + C$ So, $C = -1$. Therefore, the solution is: $y = \sqrt{2} \sqrt{1 - e^{-x}} - 1$

Step 5: Find the x-intercept

The x-intercept occurs when $y = 0$. So, we set $y = 0$ and solve for $x$: $0 = \sqrt{2} \sqrt{1 - e^{-x}} - 1$ $1 = \sqrt{2} \sqrt{1 - e^{-x}}$ $\frac{1}{\sqrt{2}} = \sqrt{1 - e^{-x}}$ $\frac{1}{2} = 1 - e^{-x}$ $e^{-x} = 1 - \frac{1}{2} = \frac{1}{2}$ $e^{-x} = \frac{1}{2}$ Taking the natural logarithm of both sides: $-x = \ln\left(\frac{1}{2}\right) = -\ln 2$ $x = \ln 2$ So, $\alpha = \ln 2$.

Step 6: Calculate $e^\alpha$

We want to find $e^\alpha$. Since $\alpha = \ln 2$, $e^\alpha = e^{\ln 2} = 2$ However, the correct answer is 1. Let's re-examine step 2:

$\frac{dy}{dx} = \frac{\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)}{\sqrt {{e^{2x}} - 1}} = \frac{\sqrt{\frac{1+e^{-x}}{2}}}{\sqrt{e^{2x}-1}} = \frac{\sqrt{1+e^{-x}}}{\sqrt{2(e^{2x}-1)}} = \frac{\sqrt{1+e^{-x}}}{\sqrt{2(e^x-1)(e^x+1)}}$ $=\frac{\sqrt{e^x+1}}{\sqrt{2e^x(e^x-1)(e^x+1)}} = \frac{1}{\sqrt{2e^x(e^x-1)}} = \frac{1}{\sqrt{2(e^{2x}-e^x)}}$ $\int dy = \int \frac{1}{\sqrt{2(e^{2x}-e^x)}}dx$ $y = \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{e^{2x}-e^x}}dx + C$ Let $u = e^{x/2}$, so $u^2 = e^x$ and $u^4=e^{2x}$. Then $du = \frac{1}{2}e^{x/2}dx = \frac{1}{2}udx$. So $dx = \frac{2du}{u}$. $y = \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{u^4-u^2}}\frac{2du}{u} + C = \frac{2}{\sqrt{2}}\int \frac{1}{u\sqrt{u^2-1}}\frac{du}{u} + C = \sqrt{2}\int \frac{du}{u^2\sqrt{u^2-1}}+C$ Let $u = \sec \theta$, $du = \sec\theta\tan\theta d\theta$. $y = \sqrt{2}\int \frac{\sec\theta\tan\theta d\theta}{\sec^2\theta \tan\theta}+C = \sqrt{2}\int \frac{d\theta}{\sec\theta}+C = \sqrt{2}\int \cos\theta d\theta+C$ $y = \sqrt{2}\sin\theta + C = \sqrt{2}\frac{\sqrt{\sec^2\theta - 1}}{\sec\theta}+C = \sqrt{2}\frac{\sqrt{u^2-1}}{u}+C = \sqrt{2}\frac{\sqrt{e^x-1}}{e^{x/2}}+C$ $y = \sqrt{2}\sqrt{\frac{e^x-1}{e^x}}+C = \sqrt{2}\sqrt{1-e^{-x}}+C$ If $x=0$, $y=-1$, so $-1 = \sqrt{2}\sqrt{1-1}+C = C$, thus $C=-1$. $y = \sqrt{2}\sqrt{1-e^{-x}}-1$ If $y=0$, $\sqrt{2}\sqrt{1-e^{-x}} = 1$ $2(1-e^{-x})=1$ $1-e^{-x}=\frac{1}{2}$ $e^{-x}=\frac{1}{2}$ $-x = \ln(1/2) = -\ln 2$ $x = \ln 2 = \alpha$ $e^\alpha = e^{\ln 2} = 2$ There is an error in the given answer.

Common Mistakes & Tips

• Be careful with algebraic manipulations, especially when dealing with square roots.
• Double-check your integration steps and the application of initial conditions.
• Consider alternative substitutions if the initial one doesn't lead to a simpler integral.

Summary

We solved the given differential equation by separating variables, using trigonometric identities and substitution to perform the integration, and applying the initial condition to find the particular solution. We then found the x-intercept by setting $y=0$ and solving for $x = \alpha$. Finally, we calculated $e^\alpha = 2$.

The final answer is incorrect. The provided "Correct Answer: 1" is wrong. My calculations consistently lead to 2. Since I must provide the correct answer, I will assume the question has an error. I have shown all the working out.

The final answer is \boxed{2}.